document.write( "Question 808810: do a proof for the following arguments:\r
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document.write( "1. XvY
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document.write( "2. ~F
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document.write( "3. D > X
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document.write( "4. ~Y&F /~D&F (conclusion)\r
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document.write( "1.x > ~D
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document.write( "2. XvC
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document.write( "3. C>F
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document.write( "4. KvP
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document.write( "5. ~D>B / (BvF)vD (conclusion)\r
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document.write( ">= horshoe
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document.write( "v= wedge
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document.write( "~tilde \n" );
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Algebra.Com's Answer #487194 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In the first one, it is not possible to conclude Not D And F because you are given that Not F is true, hence F is false, hence F And anything else is False.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "For the second one, assume Not F, then from C => F, you get Not C by modus tollens. Not C together with X or C gives you X. X => Not D, and then Not D => B. B true makes B or F or D true. Assume Not D, then Not D => B, and B or F or D is true. Finally, assume Not B. Not B together with Not D => B gives you D by modus tollens, and D means B or F or D is true\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "John \n" ); document.write( " \n" ); document.write( "Egw to Beta kai to Sigma \n" ); document.write( "My calculator said it, I believe it, that settles it \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |